Question

A current $I$ flows in a infinitely long wire with cross section in the form of a semicircular ring of radius R. The magnitude of the magnetic induction at its axis is

• A. $\frac{{\mu }_{0}I}{{\pi }^{2}R}$
• B. $\frac{{\mu }_{0}I}{2{\pi }^{2}R}$
• C. $\frac{{\mu }_{0}I}{2\pi R}$
• D. $\frac{{\mu }_{0}I}{4\pi R}$

Answer 1

Answer: (A)

In the figure we have shown the cross section (of the given wire) lying in the XY plane. The length of the wire is along the Z-axis and the current in the wire is supposed to flow along the negative Z-direction. The broad infinitely long wire can be imagined to be made of a large number of infinitely long straight wire strips, each of small width $dl$.

With reference to the figure, we have $dl=Rd\theta$.

The magnetic flux density due to the above strip is shown as $d{B}_1$ in the figure. It has an X-component $d{B}_{1}sin\theta$ and Y-component $d{B}_{1}cos\theta$. When we consider a similar strip of the same with $dl$ located symmetrically with respect to the Y-axis, we obtain a contribution $d{B}_2$ to the flux density. The flux density $d{B}_2$ has the same magnitude as $d{B}_1$. It has X-component $d{B}_{2}sin\theta$ and Y-component $d{B}_{2}cos\theta$. The X-components of $d{B}_1$ and $d{B}_1$ are in the same magnitude and direction and they add up. But the Y-components of $d{B}_1$ and $d{B}_1$ are in opposite directions and have the same magnitude. Therefore they get canceled. The entire conductor therefore produces a resultant magnetic field along the negative X-direction.

The wire strip of width $dl$ can be imagined to be an ordinary thin straight infinitely long wire carrying current $\frac{Idl}{\pi R}$ since the total current I flows through the semicircular cross section of perimeter $\pi R$. Putting $d{B}_1=d{B}_2=dB$ we have

$dB=\mu \frac{\frac{Idl}{\pi R}}{2\pi R}={\mu }_0\frac{\frac{IRd\theta }{\pi R}}{2\pi R}=\frac{{\mu }_{0}Id\theta }{2{\pi }^{2}R}$

The X-component of the above field is $\frac{{\mu }_{0}I}{2{\pi }^{2}R}sin\theta d\theta$

The field due to the entire conductor is $B={\int }_0^{\pi }\left[\frac{{\mu }_{0}I}{2{\pi }^{2}R}sin\theta \right]d\theta$

or, $B=\frac{{\mu }_{0}I}{{\pi }^{2}R}$ since ${\int }_0^{\pi }sin\theta d\theta =2$